#include<cstdio> 
#include<algorithm>
#define fo(i,a,b) for(i=a;i<=b;i++)
using namespace std;
typedef long long ll;
const int maxn=3000+10,mo=1000000007;
int f[maxn][maxn],right[maxn],sum[maxn];
int i,j,k,l,r,t,n,m,ans,one,zero;
char get(){
    char ch=getchar();
    while (ch!='0'&&ch!='1') ch=getchar();
    return ch;
}
int main(){
    scanf("%d%d",&n,&m);
    fo(i,1,n) sum[i]=get()-'0';
    fo(i,1,n+1) sum[i]+=sum[i-1];
    fo(i,1,n+1) right[i]=i;
    fo(i,1,m){
        scanf("%d%d",&j,&k);
        right[j]=max(right[j],k);
    }
    fo(i,1,n) right[i]=max(right[i],right[i-1]);
    f[1][sum[right[1]]]=1;
    fo(i,1,n)
        fo(j,0,n)
            if (f[i][j]){
                l=right[i]+1;r=right[i+1];
                one=j;zero=right[i]-i+1-j;
                if (zero) (f[i+1][j+sum[r]-sum[l-1]]+=f[i][j])%=mo;
                if (one) (f[i+1][j-1+sum[r]-sum[l-1]]+=f[i][j])%=mo;
            }
    ans=f[n+1][0];
    (ans+=mo)%=mo;
    printf("%d\n",ans);
}
### Prompt

Your task is to create a Cpp solution to the following 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

### Response

```cpp
#include<cstdio> 
#include<algorithm>
#define fo(i,a,b) for(i=a;i<=b;i++)
using namespace std;
typedef long long ll;
const int maxn=3000+10,mo=1000000007;
int f[maxn][maxn],right[maxn],sum[maxn];
int i,j,k,l,r,t,n,m,ans,one,zero;
char get(){
    char ch=getchar();
    while (ch!='0'&&ch!='1') ch=getchar();
    return ch;
}
int main(){
    scanf("%d%d",&n,&m);
    fo(i,1,n) sum[i]=get()-'0';
    fo(i,1,n+1) sum[i]+=sum[i-1];
    fo(i,1,n+1) right[i]=i;
    fo(i,1,m){
        scanf("%d%d",&j,&k);
        right[j]=max(right[j],k);
    }
    fo(i,1,n) right[i]=max(right[i],right[i-1]);
    f[1][sum[right[1]]]=1;
    fo(i,1,n)
        fo(j,0,n)
            if (f[i][j]){
                l=right[i]+1;r=right[i+1];
                one=j;zero=right[i]-i+1-j;
                if (zero) (f[i+1][j+sum[r]-sum[l-1]]+=f[i][j])%=mo;
                if (one) (f[i+1][j-1+sum[r]-sum[l-1]]+=f[i][j])%=mo;
            }
    ans=f[n+1][0];
    (ans+=mo)%=mo;
    printf("%d\n",ans);
}
```