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>
using namespace std;
#define N 200100
#define INF 0x3f3f3f3f

int n,c0,q,x,y;
int b[N],cost[N],d[N<<2],dp[N];
vector<int> v[N];

int query(int o,int L,int R,int l,int r) {
	if (l<=L && r>=R) return d[o];
	int mid=(L+R)>>1,ret=INF;
	if (l<=mid) ret=min(ret,query(o<<1,L,mid,l,r));
	if (r>mid) ret=min(ret,query(o<<1|1,mid+1,R,l,r));
	return ret;
}

void upd(int o,int l,int r,int x,int y) {
	if (l==r) {d[o]=min(d[o],y); return;}
	int mid=(l+r)>>1;
	if (x<=mid) upd(o<<1,l,mid,x,y);
	else upd(o<<1|1,mid+1,r,x,y);
	d[o]=min(d[o<<1],d[o<<1|1]);
}

int main()
{
	scanf("%d",&n);
	for (int i=1;i<=n;i++)
	{
		scanf("%d",&b[i]);
		if (!b[i]) c0++;
		cost[i]=b[i] ? 1 : -1;
	}
	scanf("%d",&q);
	for (int i=1;i<=q;i++)
	{
		scanf("%d%d",&x,&y);
		v[x].push_back(y);
	}
	memset(d,0x3f,sizeof(d));
	memset(dp,0x3f,sizeof(dp));
	dp[0]=0;
	for (int i=1;i<=n;i++)
	{
		for (int j=0;j<(int)v[i].size();j++)
		{
			int x=v[i][j];
			int t=min(dp[i-1],query(1,1,n,max(i-1,1),x));
			if (t<dp[x]) {
				dp[x]=t;
				upd(1,1,n,x,t);
			}
		}
		dp[i]=min(dp[i],dp[i-1]+cost[i]);
	}
	printf("%d\n",dp[n]+c0);
	return 0;
}