Problem: There is a string S of length N consisting of characters `0` and `1`. You will perform the following operation for each i = 1, 2, ..., m:

* Arbitrarily permute the characters within the substring of S starting at the l_i-th character from the left and extending through the r_i-th character.



Here, the sequence l_i is non-decreasing.

How many values are possible for S after the M operations, modulo 1000000007(= 10^9+7)?

Constraints

* 2≦N≦3000
* 1≦M≦3000
* S consists of characters `0` and `1`.
* The length of S equals N.
* 1≦l_i < r_i≦N
* l_i ≦ l_{i+1}

Input

The input is given from Standard Input in the following format:


N M
S
l_1 r_1
:
l_M r_M


Output

Print the number of the possible values for S after the M operations, modulo 1000000007.

Examples

Input

5 2
01001
2 4
3 5


Output

6


Input

9 3
110111110
1 4
4 6
6 9


Output

26


Input

11 6
00101000110
2 4
2 3
4 7
5 6
6 10
10 11


Output

143

Solution (in CPP):

Solution:
#include <cctype>
#include <cstdio>
#include <algorithm>
#define rep(i,x,y) for (int i=x; i<=y; ++i)
#define repd(i,x,y) for (int i=x; i>=y; --i)

int get()
{
    char c;
    while (!isdigit(c=getchar()));
    int k=c-'0';
    for (; isdigit(c=getchar()); k=k*10+c-'0');
    return k;
}

using namespace std;
const int N=3010,mod=1e9+7;
char s[N];
int n,m,ans,c[N][N],f[N][N],a[N];
struct data
{
    int l,r;
    bool operator<(const data &x) const
    {
        return l<x.l || (l==x.l && r>x.r);
    }
} b[N];

int main()
{
    n=get(),m=get();
    rep(i,0,n)
    {
        c[i][0]=1;
        rep(j,1,i)
            c[i][j]=(c[i-1][j-1]+c[i-1][j])%mod;
    }
    scanf("%s",s+1);
    rep(i,1,n)
        a[i]=a[i-1]+s[i]-48;
    rep(i,1,m)
        b[i].l=get(),b[i].r=get();
    sort(b+1,b+1+m);
    int t=0;
    rep(i,1,m)
        if (b[i].r>b[t].r)
            b[++t]=b[i];
    m=t,b[m+1].l=n+1,f[0][0]=1;
    rep(i,1,m)
    {
        int w=a[b[i].r]-a[max(b[i].l-1,b[i-1].r)];
        int l=min(b[i].r,b[i+1].l-1)-b[i].l+1;
        int mx=b[i].r-b[i].l+1-l;
        rep(j,0,l)
            rep(k,max(0,w-j),mx)
                f[i][k]=(f[i][k]+1ll*f[i-1][j+k-w]*c[l][j])%mod;
    }
    rep(i,0,n)
        ans=(ans+f[m][i])%mod;
    printf("%d",ans);
    return 0;
}