Problem: You are given an integer sequence of length N: A_1,A_2,...,A_N. Let us perform Q operations in order. The i-th operation is described by two integers X_i and Y_i. In this operation, we will choose one of the following two actions and perform it:

* Swap the values of A_{X_i} and A_{Y_i}
* Do nothing



There are 2^Q ways to perform these operations. Find the sum of the inversion numbers of the final sequences for all of these ways to perform operations, modulo 10^9+7.

Here, the inversion number of a sequence P_1,P_2,...,P_M is the number of pairs of integers (i,j) such that 1\leq i < j\leq M and P_i > P_j.

Constraints

* 1 \leq N \leq 3000
* 0 \leq Q \leq 3000
* 0 \leq A_i \leq 10^9(1\leq i\leq N)
* 1 \leq X_i,Y_i \leq N(1\leq i\leq Q)
* X_i\neq Y_i(1\leq i\leq Q)
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N Q
A_1
:
A_N
X_1 Y_1
:
X_Q Y_Q


Output

Print the sum of the inversion numbers of the final sequences, modulo 10^9+7.

Examples

Input

3 2
1
2
3
1 2
1 3


Output

6


Input

5 3
3
2
3
1
4
1 5
2 3
4 2


Output

36


Input

9 5
3
1
4
1
5
9
2
6
5
3 5
8 9
7 9
3 2
3 8


Output

425

Solution (in CPP):

Solution:
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <stack>
#include <queue>
#define rep(i,l,r) for(int i=(l);i<=(r);++i)
#define per(i,r,l) for(int i=(r);i>=(l);--i)
using namespace std;
const int P=1e9+7,N=3010,i2=(P+1)>>1;
int n,Q,a[N],dp[N][N],tmp[N][N],ans,tot=1;
int main(){
    scanf("%d%d",&n,&Q);
    rep(i,1,n) scanf("%d",a+i);
    rep(i,1,n) rep(j,1,n) dp[i][j]=a[i]>a[j];
    while(Q--){
        int x,y;scanf("%d%d",&x,&y);
        rep(i,1,n) if((i^x)&&(i^y)) tmp[i][x]=dp[i][y],tmp[x][i]=dp[y][i],tmp[i][y]=dp[i][x],tmp[y][i]=dp[x][i];
        rep(i,1,n) if((i^x)&&(i^y))
            dp[i][x]=1ll*i2*(dp[i][x]+tmp[i][x])%P,
            dp[i][y]=1ll*i2*(dp[i][y]+tmp[i][y])%P,
            dp[x][i]=1ll*i2*(dp[x][i]+tmp[x][i])%P,
            dp[y][i]=1ll*i2*(dp[y][i]+tmp[y][i])%P;
        rep(i,1,n) if((i^x)&&(i^y)) tmp[i][x]=tmp[i][y]=tmp[x][i]=tmp[y][i]=0;
        dp[x][y]=dp[y][x]=1ll*i2*(dp[x][y]+dp[y][x])%P;
        tot=2ll*tot%P;
    }
    rep(i,1,n-1) rep(j,i+1,n) ans=(ans+dp[i][j])%P;
    printf("%lld\n",1ll*ans*tot%P);
    return 0;
}