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<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
using namespace std;
const int N=200005;
vector<int> vct[N];
int n,q,i,j,b[N],mn[N*5],add[N*5],rt[N],ans;
void built(int o,int l,int r)
{
	if(l==r)
	{
		if(l!=0)
			mn[o]=1<<30;
		else
			mn[o]=0;
		return;
	}
	int mid=l+r>>1;
	built(o<<1,l,mid);
	built(o<<1|1,mid+1,r);
	mn[o]=min(mn[o<<1],mn[o<<1|1]);
}
void pushdown(int o,int l,int r)
{
	if(add[o])
	{
		add[o<<1]+=add[o],add[o<<1|1]+=add[o];
		mn[o<<1]+=add[o],mn[o<<1|1]+=add[o];
		add[o]=0;
	}
}
void update2(int o,int l,int r,int x,int y,int z)
{
	if(l>=x&&r<=y)
	{
		mn[o]+=z;
		add[o]+=z;
		return;
	}
	pushdown(o,l,r);
	int mid=l+r>>1;
	if(x<=mid)
		update2(o<<1,l,mid,x,y,z);
	if(y>mid)
		update2(o<<1|1,mid+1,r,x,y,z);
	mn[o]=min(mn[o<<1],mn[o<<1|1]);
}
void update(int o,int l,int r,int x,int y)
{
	if(l==r)
	{
		mn[o]=min(mn[o],y);
		return;
	}
	pushdown(o,l,r);
	int mid=l+r>>1;
	if(x<=mid)
		update(o<<1,l,mid,x,y);
	else
		update(o<<1|1,mid+1,r,x,y);
	mn[o]=min(mn[o<<1],mn[o<<1|1]);
}
int query(int o,int l,int r,int x,int y)
{
	if(l>=x&&r<=y)
		return mn[o];
	pushdown(o,l,r);
	int mid=l+r>>1,rtn=1<<30;
	if(x<=mid)
		rtn=query(o<<1,l,mid,x,y);
	if(y>mid)
		rtn=min(rtn,query(o<<1|1,mid+1,r,x,y));
	return rtn;
}
int main()
{
	scanf("%d",&n);
	for(i=1;i<=n;++i)
	{
		scanf("%d",b+i);
		if(b[i]==0)
			++ans;
	}
	built(1,0,n);
	scanf("%d",&q);
	while(q--)
	{
		scanf("%d%d",&i,&j);
		vct[i].push_back(j);
	}
	for(i=1;i<=n;++i)
	{
		for(j=vct[i].size()-1;j>=0;--j)
			update(1,0,n,vct[i][j],query(1,0,n,0,vct[i][j]-1));
		if(b[i]==1)
			update2(1,0,n,0,i-1,1);
		else
			update2(1,0,n,0,i-1,-1);
	}
	ans+=query(1,0,n,0,n);
	printf("%d",ans);
	return 0;
}