polycarp loves ciphers. He has invented his own cipher called Right-Left.

Right-Left cipher is used for strings. To encrypt the string s=s1s2…sns=s1s2…sn polycarp uses the following algorithm:

• he writes down s1s1,
• he appends the current word with s2s2 (i.e. writes down s2s2 to the right of the current result),
• he prepends the current word with s3s3 (i.e. writes down s3s3 to the left of the current result),
• he appends the current word with s4s4 (i.e. writes down s4s4 to the right of the current result),
• he prepends the current word with s5s5 (i.e. writes down s5s5 to the left of the current result),
• and so on for each position until the end of ss.

For example,if ss="techno" the process is: "t" →→ "te" →→ "cte" →→ "cteh" →→ "ncteh" @H_797_301@→→ "ncteho". So the encrypted ss="techno" is "ncteho".

Given string tt — the result of encryption of some string ss. Your task is to decrypt it,i.e. find the string ss.

The only line of the input contains tt — the result of encryption of some string ss. It contains only lowercase Latin letters. The length of tt is between 11 and 5050,inclusive.

Print such string ss that after encryption it equals tt.

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;

LL xx[105],yy[105],cnt;

LL gcd(LL x,LL y){
    int res=(y==0?x:gcd(y,x%y));
    xx[++cnt]=x;
    yy[cnt]=y;
    return res;
}

LL ans[105];
bool check(LL a,LL b){
    if(a%2==1){
        return (b/a-1)%2;
    }
    else{
        if(a>=1000000000||b<a*a){
            return !((b/a)%2);
        }
        else{
            LL res=b/a;
            LL c1=(res%(a+1))%2;
            LL c2=res%(a+1);
            return !c1||(c2==a);
        }
    }
}
void solve(LL a,LL b){
    cnt=0;
    gcd(a,b);
    ans[1]=0;
    for(int i=2;i<=cnt;i++){
        if(ans[i-1]==0) ans[i]=1;
        else if(check(yy[i],xx[i]-yy[i-1])) ans[i]=1;
        else ans[i]=0;
    }
    if(ans[cnt]) printf("First\n");
    else printf("Second\n");
}

int p=12;
int biao[100005];
int dabiao(int x){
    if(biao[x]!=-1) return biao[x];
    if(x<0) return 0;
    int temp=p;
    int flag=0;
    while(temp<=x*p){
        if(dabiao(x-temp/p)==0){
            flag=1; break;
        }
        temp*=p;
    }
    return biao[x]=flag;
}

int main(){
    /*******
    memset(biao,-1,sizeof(biao));
    biao[0]=1;
    dabiao(100);
    for(int i=1;i<=100;i++){
        cout<<i<<" "<<biao[i]<<endl;
    }
    *******/
    int n;
    cin>>n;
    while(n--){
        LL a,b;
        cin>>a>>b;
        if(a<b) swap(a,b);
        if(b==0) cout<<"Second"<<endl;
        else if(a%b==0) cout<<"First"<<endl;
        else solve(a,b);
    }

    return 0;
}