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>
using namespace std;
const int N=3005,Mod=1e9+7,iM=Mod+1>>1;
int n,q,a[N],f[N][N],t[N][N],ans,p=1;
#define mul(x,y) (1ll*(x)*(y)%Mod)
inline int add(int x, int y)
{	return (x+y>=Mod?x+y-Mod:x+y);
}
int main()
{
	scanf("%d%d",&n,&q);
	for(int i=1;i<=n;++i) scanf("%d",&a[i]);
	for(int i=1;i<=n;++i)
		for(int j=1;j<=n;++j) f[i][j]=(a[i]>a[j]);
	while(q--)
	{
		int x,y;
		scanf("%d%d",&x,&y);
		if(x>y) swap(x,y);
		if(x==y) continue;
		t[x][y]=mul(iM,add(f[x][y],f[y][x]));
		for(int i=1;i<=n;++i) if(i!=x&&i!=y)
			t[x][i]=mul(iM,add(f[x][i],f[y][i])),
			t[i][x]=mul(iM,add(f[i][x],f[i][y])),
			t[y][i]=mul(iM,add(f[y][i],f[x][i])),
			t[i][y]=mul(iM,add(f[i][y],f[i][x]));
		f[x][y]=f[y][x]=t[x][y];
		for(int i=1;i<=n;++i) if(i!=x&&i!=y)
			f[x][i]=t[x][i],f[i][x]=t[i][x],f[y][i]=t[y][i],f[i][y]=t[i][y];
		p=mul(2,p);
	}
	for(int i=1;i<=n;++i)
		for(int j=i+1;j<=n;++j) ans=add(ans,f[i][j]);
	printf("%d\n",mul(ans,p));
}