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:
//ΔAGC030D
#include<iostream>
#include<cstdio>
#include<fstream>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<bitset>
#include<cmath>
#include<cstring>
#include<cstdlib>
using namespace std;
typedef long long LL;
typedef double DB;
const int N = 3333;
const int MO = 1e9+7;
int mul(int x,int y){
	return (LL)x*y%MO;
}
int n,q,a[N],f[N][N],g[N][N];
int main()
{
	int i,j,x,y,z=1,e=(MO+1)/2;
	scanf("%d%d",&n,&q);
	for(i=1;i<=n;i=i+1)
		scanf("%d",a+i);
	for(i=1;i<=n;i=i+1)
		for(j=1;j<=n;j=j+1)
			f[i][j]=a[i]>a[j];
	while(q--){
		scanf("%d%d",&x,&y);
		for(i=1;i<=n;i=i+1){
			g[i][x]=f[i][x]+f[i][y];
			g[i][y]=f[i][y]+f[i][x];
			g[x][i]=f[x][i]+f[y][i];
			g[y][i]=f[y][i]+f[x][i];
		}
		g[x][y]=f[y][x]+f[x][y],g[y][x]=f[x][y]+f[y][x];
		for(i=1;i<=n;i=i+1){
			f[i][x]=mul(g[i][x],e);
			f[i][y]=mul(g[i][y],e);
			f[x][i]=mul(g[x][i],e);
			f[y][i]=mul(g[y][i],e);
		}
		for(i=1;i<=n;i=i+1){
			g[i][x]=0;
			g[i][y]=0;
			g[x][i]=0;
			g[y][i]=0;
		}
		z=z*2%MO;
	}
	x=0;
	for(i=1;i<=n;i=i+1)
		for(j=i+1;j<=n;j=j+1)
			x=(x+f[i][j])%MO;
	cout<<mul(x,z);
	return 0;
}