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<bits/stdc++.h>
#define ll long long
#define ull unsigned ll
#define uint unsigned
#define pii pair<int,int>
#define pll pair<ll,ll>
#define PB push_back
#define fi first
#define se second
#define For(i,j,k) for (int i=(int)(j);i<=(int)(k);i++)
#define Rep(i,j,k) for (int i=(int)(j);i>=(int)(k);i--)
#define CLR(a,v) memset(a,v,sizeof(a));
#define CPY(a,b) memcpy(a,b,sizeof(a));
using namespace std;
const int N=3005;
const int mo=1000000007;
int n,Q,ans,a[N];
int f[N][N];
int main(){
	scanf("%d%d",&n,&Q);
	For(i,1,n) scanf("%d",&a[i]);
	For(i,1,n) For(j,1,n) f[i][j]=(a[i]>a[j]);
	For(i,1,Q){
		int x,y;
		scanf("%d%d",&x,&y);
		For(j,1,n) if (j!=x&&j!=y){
			f[x][j]=f[y][j]=1ll*(f[x][j]+f[y][j])*(mo+1)/2%mo;
			f[j][x]=f[j][y]=1ll*(f[j][x]+f[j][y])*(mo+1)/2%mo;
		}
		f[x][y]=f[y][x]=
			1ll*(f[x][y]+f[y][x])*(mo+1)/2%mo;
	}
	For(i,1,n) For(j,i+1,n)
		ans=(ans+f[i][j])%mo;
	For(i,1,Q) ans=2ll*ans%mo;
	printf("%d\n",ans);
}