首先判断能否构成回文串 对于前 ⌊n/2⌋个字符，若每种字符出现次数恰好为对应的一半， 那么就把右半部分对称成左半部分的样子即可，若左边相邻两个字符交换， 逆序对个数只会 ±1，而右半部分的逆序对只会 ∓1，恰好抵消 否则贪心的将右边靠左且多的字符往左边移，这样影响的逆序对数是最少的 然后树状数组求逆序对即可

#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define fi first
#define se second
#define pii pair<int, int>
#define lp p << 1
#define rp p << 1 | 1
#define mid ((l + r) >> 1)
#define ll long long
#define db double
#define rep(i,a,b) for(int i=a;i<b;i++)
#define per(i,a,b) for(int i=b-1;i>=a;i--)
#define Edg int cnt=1,head[N],to[N*2],ne[N*2];void addd(int u,int v){to[++cnt]=v;ne[cnt]=head[u];head[u]=cnt;}void add(int u,int v){addd(u,v);addd(v,u);}
#define Edgc int cnt=1,head[N],to[N*2],ne[N*2],c[N*2];void addd(int u,int v,int w){to[++cnt]=v;ne[cnt]=head[u];c[cnt]=w;head[u]=cnt;}void add(int u,int v,int w){addd(u,v,w);addd(v,u,w);}
#define es(u,i,v) for(int i=head[u],v=to[i];i;i=ne[i],v=to[i])

const int N = 1e5 + 7;
char s[N];
int cnt[28], pos[28][N], half[28], a[N], b[N], n, tree[N];
inline int lowbit(int x) { return x & -x; }
void add(int x) {
	assert(x);
	for (int i = x; i <= n; i += lowbit(i)) tree[i]++;
}
int query(int x) {
	int ans = 0;
	for (int i = x; i; i -= lowbit(i)) 
		ans += tree[i]; 
	return ans;
}

void solve() {
	memset(cnt, 0, sizeof(cnt));
	scanf("%s", s + 1);
	n = strlen(s + 1);
	memset(tree, 0, sizeof(int) * (n + 2));
	rep (i, 1, n + 1) {
		int c = s[i] - 'a';
		cnt[c]++;
		pos[c][cnt[c]] = i;
	}
	int tol = 0, mid_alp = -1;
	rep (i, 0, 26) {
		if (cnt[i] & 1) tol++, mid_alp = i;
		half[i] = cnt[i] / 2;
	}
	if (n % 2 == 0) {
		if (tol) { puts("Impossible"); return; }
	} else if (tol != 1) {
		puts("Impossible"); return;
	}
	int cur = 1;
	int mi = n / 2;
	rep (i, 1, mi + 1) {
		while (!half[s[cur] - 'a']) cur++;
		a[i] = cur; half[s[cur++] - 'a']--;
	}
	if (n & 1) {
		a[mi + 1] = pos[mid_alp][cnt[mid_alp] / 2 + 1];
		//assert(pos[mid_alp][cnt[mid_alp] / 2 + 1] == cnt[mid_alp] / 2 + 1);
		per(i, 1, mi + 1) {
			int c = s[a[mi - i + 1]] - 'a';
			a[i + mi + 1] = pos[c][cnt[c]--];
		}
	} else {
		per(i, 1, mi + 1) {
			int c = s[a[mi - i + 1]] - 'a';
			a[i + mi] = pos[c][cnt[c]--];
		}
	}
	rep (i, 1, n + 1) b[a[i]] = i;
	long long ans = 0;
	per (i, 1, n + 1) ans += query(b[i]), add(b[i]);
	printf("%lld\n", ans);
}

int main() {
	int T;
	scanf("%d", &T);
	while (T--)
		solve();
	return 0;
}

/*
首先判断能否构成回文串
对于前 ⌊n/2⌋个字符，若每种字符出现次数恰好为对应的一半，
那么就把右半部分对称成左半部分的样子即可，若左边相邻两个字符交换，
逆序对个数只会 ±1，而右半部分的逆序对只会 ∓1，恰好抵消
否则贪心的将右边靠左且多的字符往左边移，这样影响的逆序对数是最少的
然后树状数组求逆序对即可
*/
#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define fi first
#define se second
#define pii pair<int, int>
#define lp p << 1
#define rp p << 1 | 1
#define mid ((l + r) >> 1)
#define ll long long
#define db double
#define rep(i,a,b) for(int i=a;i<b;i++)
#define per(i,a,b) for(int i=b-1;i>=a;i--)
#define Edg int cnt=1,head[N],to[N*2],ne[N*2];void addd(int u,int v){to[++cnt]=v;ne[cnt]=head[u];head[u]=cnt;}void add(int u,int v){addd(u,v);addd(v,u);}
#define Edgc int cnt=1,head[N],to[N*2],ne[N*2],c[N*2];void addd(int u,int v,int w){to[++cnt]=v;ne[cnt]=head[u];c[cnt]=w;head[u]=cnt;}void add(int u,int v,int w){addd(u,v,w);addd(v,u,w);}
#define es(u,i,v) for(int i=head[u],v=to[i];i;i=ne[i],v=to[i])

const int N = 1e5 + 7;
char s[N];
int cnt[28], pos[28][N], half[28], a[N], b[N], n, tree[N];
inline int lowbit(int x) { return x & -x; }
void add(int x) {
	assert(x);
	for (int i = x; i <= n; i += lowbit(i)) tree[i]++;
}
int query(int x) {
	int ans = 0;
	for (int i = x; i; i -= lowbit(i)) 
		ans += tree[i]; 
	return ans;
}

void solve() {
	memset(cnt, 0, sizeof(cnt));
	scanf("%s", s + 1);
	n = strlen(s + 1);
	memset(tree, 0, sizeof(int) * (n + 2));
	rep (i, 1, n + 1) {
		int c = s[i] - 'a';
		cnt[c]++;
		pos[c][cnt[c]] = i;
	}
	int tol = 0, mid_alp = -1;
	rep (i, 0, 26) {
		if (cnt[i] & 1) tol++, mid_alp = i;
		half[i] = cnt[i] / 2;
	}
	if (n % 2 == 0) {
		if (tol) { puts("Impossible"); return; }
	} else if (tol != 1) {
		puts("Impossible"); return;
	}
	int cur = 1;
	int mi = n / 2;
	rep (i, 1, mi + 1) {
		while (!half[s[cur] - 'a']) cur++;
		a[i] = cur; half[s[cur++] - 'a']--;
	}
	if (n & 1) {
		a[mi + 1] = pos[mid_alp][cnt[mid_alp] / 2 + 1];
		//assert(pos[mid_alp][cnt[mid_alp] / 2 + 1] == cnt[mid_alp] / 2 + 1);
		per(i, 1, mi + 1) {
			int c = s[a[mi - i + 1]] - 'a';
			a[i + mi + 1] = pos[c][cnt[c]--];
		}
	} else {
		per(i, 1, mi + 1) {
			int c = s[a[mi - i + 1]] - 'a';
			a[i + mi] = pos[c][cnt[c]--];
		}
	}
	rep (i, 1, n + 1) b[a[i]] = i;
	long long ans = 0;
	per (i, 1, n + 1) ans += query(b[i]), add(b[i]);
	printf("%lld\n", ans);
}

int main() {
	int T;
	scanf("%d", &T);
	while (T--)
		solve();
	return 0;
}