Problem: You are given a sequence a = \\{a_1, ..., a_N\\} with all zeros, and a sequence b = \\{b_1, ..., b_N\\} consisting of 0 and 1. The length of both is N.

You can perform Q kinds of operations. The i-th operation is as follows:

* Replace each of a_{l_i}, a_{l_i + 1}, ..., a_{r_i} with 1.



Minimize the hamming distance between a and b, that is, the number of i such that a_i \neq b_i, by performing some of the Q operations.

Constraints

* 1 \leq N \leq 200,000
* b consists of 0 and 1.
* 1 \leq Q \leq 200,000
* 1 \leq l_i \leq r_i \leq N
* If i \neq j, either l_i \neq l_j or r_i \neq r_j.

Input

Input is given from Standard Input in the following format:


N
b_1 b_2 ... b_N
Q
l_1 r_1
l_2 r_2
:
l_Q r_Q


Output

Print the minimum possible hamming distance.

Examples

Input

3
1 0 1
1
1 3


Output

1


Input

3
1 0 1
2
1 1
3 3


Output

0


Input

3
1 0 1
2
1 1
2 3


Output

1


Input

5
0 1 0 1 0
1
1 5


Output

2


Input

9
0 1 0 1 1 1 0 1 0
3
1 4
5 8
6 7


Output

3


Input

15
1 1 0 0 0 0 0 0 1 0 1 1 1 0 0
9
4 10
13 14
1 7
4 14
9 11
2 6
7 8
3 12
7 13


Output

5


Input

10
0 0 0 1 0 0 1 1 1 0
7
1 4
2 5
1 3
6 7
9 9
1 5
7 9


Output

1

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
#define ll long long
using namespace std;
int n,m,ans,mlby,f[1000010],a[500010],b[500010];
vector<int>zh[200010];
/*void build(int x,int l,int r){
	f[x]=-1e9;
	if(l==r)return;
	int mid=()
}*/
void ins(int x,int l,int r,int t,int w){
	if(l==r){
		f[x]=max(f[x],w);return;
	}
	int mid=(l+r)/2;
	if(mid>=t)ins(x*2,l,mid,t,w);
	 else ins(x*2+1,mid+1,r,t,w);
	f[x]=max(f[x*2],f[x*2+1]);
}
void xiu(int x,int l,int r,int t,int w){
	if(l==r){
		f[x]=w;return;
	}
	int mid=(l+r)/2;
	if(mid>=t)xiu(x*2,l,mid,t,w);
	 else xiu(x*2+1,mid+1,r,t,w);
	f[x]=max(f[x*2],f[x*2+1]);
}
int query(int x,int l,int r,int t,int k){
	if(t>k)return -1e9;
	if(l==t&&r==k)return f[x];
	int mid=(l+r)/2;
	if(mid>=k)return query(x*2,l,mid,t,k);
	 else if(t>mid)return query(x*2+1,mid+1,r,t,k);
	  else return max(query(x*2,l,mid,t,mid),query(x*2+1,mid+1,r,mid+1,k));
}
int main(){
	int i,j,x,y,xlh;
	scanf("%d",&n);
	for(i=1;i<=n;i++)scanf("%d",&b[i]);
	for(i=1;i<=n;i++){
		ans+=(b[i]==0);
		if(b[i]==1)a[i]=1;
		 else a[i]=-1;
	}
	for(i=1;i<=n;i++)a[i]+=a[i-1];
	scanf("%d",&m);
	for(i=1;i<=m;i++){
		scanf("%d%d",&x,&y);
		zh[x].push_back(y);
	}
	memset(f,172,sizeof(f));
	//build(1,1,n);
	for(i=1;i<=n;i++){
		for(j=0;j<zh[i].size();j++){
			x=zh[i][j];
			mlby=max(mlby,a[x]-a[i-1]);
			ins(1,1,n,x,a[x]-a[i-1]-a[x]);
			xlh=max(query(1,1,n,1,i-1)-a[i-1],query(1,1,n,i,x));
			mlby=max(mlby,xlh+a[x]);
			ins(1,1,n,x,xlh);
		}
		xlh=query(1,1,n,i,i);
		//printf("%d %d\n",i,xlh+a[i]);
		xiu(1,1,n,i,xlh+a[i]);
	}
	printf("%d",n-(mlby+ans));
}