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<iostream>
using namespace std;
const int mod=1e9+7;
const int r=(mod+1)>>1;
inline int add(int x)
{
	return x>=mod?x-mod:x;
}
int n,Q;
int a[3050];
int f[3050][3050]; 
int t[3050][3050];
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)
		{
			if(a[i]>a[j]) f[i][j]=1;
		}
	} 
	for(int i=1;i<=Q;++i)
	{
		int x,y; scanf("%d%d",&x,&y);
		for(int j=1;j<=n;++j) if(j!=x)
		{
			t[j][x]=1ll*(f[j][x]+f[j][y])*r%mod;
			t[x][j]=1ll*(f[x][j]+f[y][j])*r%mod;
		}
		for(int j=1;j<=n;++j) if(j!=y)
		{
			t[j][y]=1ll*(f[j][y]+f[j][x])*r%mod;
			t[y][j]=1ll*(f[y][j]+f[x][j])*r%mod;
		}
		t[x][y]=1ll*(f[x][y]+f[y][x])*r%mod;
		t[y][x]=1ll*(f[y][x]+f[x][y])*r%mod;
		for(int j=1;j<=n;++j) if(j!=x)
		{
			f[j][x]=t[j][x];
			f[x][j]=t[x][j];
		}
		for(int j=1;j<=n;++j) if(j!=y)
		{
			f[j][y]=t[j][y];
			f[y][j]=t[y][j];
		}
		f[x][y]=t[x][y];
		f[y][x]=t[y][x];
	}
	int an=0;
	for(int i=1;i<=n;++i)
	{
		for(int j=i+1;j<=n;++j)
		{
			an=add(an+f[i][j]);
		}
	}
	for(int i=1;i<=Q;++i)
	{
		an=add(an<<1);
	}
	printf("%d\n",an);
	return 0;
}