分析：

不算太難的fail指標應用題。每次利用fail指標在上一個位置的匹配資訊，不停地向下匹配，如果超過一半則退回去即可。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<map>
#define SF scanf
#define PF printf
#define MAXN 1000010
#define MOD 1000000007
using namespace std;
typedef long long ll;
int n;
char s[MAXN]
 
;
int fail[MAXN],dep[MAXN];
ll ans;
void build(){
	int las=-1;
	fail[0]=-1;
	for(int i=0;i<n;i++){
		while(las!=-1&&s[las]!=s[i])
			las=fail[las];
		fail[i+1]=las+1;
		dep[i+1]=dep[fail[i+1]]+1;
		//PF("[%d %d %d]\n",i+1,fail[i+1],dep[i+1]);
		las++;
	}
}
void solve(){
	int las=-1;
	for(int i=
 
0;i<n;i++){
		while(las!=-1&&s[i]!=s[las])
			las=fail[las];
		las++;
		while(las!=-1&&(las<<1)>i+1)
			las=fail[las];
		//PF("---[%d %lld %lld]---\n",las,ans,dep[las]+1ll);
		if(las!=-1)
			ans=ans*1ll*(dep[las]+1)%MOD;
	}
}
int main(){
	int t;
	SF("%d",&t);
	while(t--)
 
{
		SF("%s",s);
		n=strlen(s);
		memset(fail,0,sizeof fail);
		memset(dep,0,sizeof dep);
		build();
		ans=1;
		solve();
		PF("%lld\n",ans);
	}
}