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 <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;
const int N=3010,Mo=1000000007,P=(Mo+1)/2;
int a[N],f[N][N],g[N][N];
inline int gi() {
    int x=0,o=1;
    char ch=getchar();
    while(ch<'0'||ch>'9') ch=='-'?o=-1:0,ch=getchar();
    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return o*x;
}
int main() {
    int n,Q,ans=0,t=1;
    cin>>n>>Q;
    for(int i=1;i<=n;i++) a[i]=gi();
    for(int i=1;i<=n;i++)
	for(int j=1;j<=n;j++)
	    f[i][j]=(a[i]<a[j]),g[i][j]=(a[i]>a[j]);
    for(int i=1;i<=Q;i++) {
	int x=gi(),y=gi();t=2*t%Mo;
	for(int j=1;j<=n;j++)
	    if(j!=x&&j!=y) {
		g[x][j]=g[y][j]=f[j][x]=f[j][y]=1LL*P*(f[j][x]+f[j][y])%Mo;
		f[x][j]=f[y][j]=g[j][x]=g[j][y]=1LL*P*(g[j][x]+g[j][y])%Mo;
	    }
	f[x][y]=g[x][y]=f[y][x]=g[y][x]=1LL*P*(f[x][y]+g[x][y])%Mo;
    }
    for(int i=1;i<=n;i++)
	for(int j=i+1;j<=n;j++)
	    ans=(ans+1LL*g[i][j]*t)%Mo;
    cout<<ans;
    return 0;
}