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;

inline int read()
{
	int x=0;char ch=getchar();
	while(ch<'0' || '9'<ch)ch=getchar();
	while('0'<=ch && ch<='9')x=x*10+(ch^48),ch=getchar();
	return x;
}

inline void chkmin(int &a,int b){if(a>b)a=b;}
inline int minn(int a,int b){if(a<b)return a;return b;}

const int N=200009;
const int Q=200009;
const int Inf=2e9+7;

struct node
{
	int l,r;
}qu[Q];

int n,q,ans;
int b[N],sum[N];
int f[N],t[N<<2];
vector<int> g[N];

inline int zero(int l,int r)
{
	return r-(l-1)-(sum[r]-sum[l-1]);
}

inline int one(int l,int r)
{
	if(l>r)return 0;
	return sum[r]-sum[l-1];
}

#define mid ((l+r)>>1)

inline void upd(int z)
{
	t[z]=minn(t[z<<1],t[z<<1|1]);
}

inline void build(int x,int l,int r)
{
	if(l==r)
	{
		t[x]=f[l]-sum[l];
		return;
	}
	build(x<<1,l,mid);
	build(x<<1|1,mid+1,r);
	upd(x);
}

inline void modify(int x,int l,int r,int p)
{
	if(l==r)
	{
		t[x]=f[l]-sum[l];
		return;
	}
	if(p<=mid)
		modify(x<<1,l,mid,p);
	else 
		modify(x<<1|1,mid+1,r,p);
	upd(x);
}

inline int query(int x,int l,int r,int dl,int dr)
{
	if(dl==l && r==dr)
		return t[x];
	if(dr<=mid)
		return query(x<<1,l,mid,dl,dr);
	else if(mid<dl)
		return query(x<<1|1,mid+1,r,dl,dr);
	return minn(query(x<<1,l,mid,dl,mid),query(x<<1|1,mid+1,r,mid+1,dr));
}

int main()
{
	if(fopen("hamming.in","r"))
	{
		freopen("hamming.in","r",stdin);
		freopen("hamming.out","w",stdout);
	}

	n=read();
	for(int i=1;i<=n;i++)
	{
		b[i]=read()^1;
		sum[i]=sum[i-1]+b[i];
	}
	q=read();
	for(int i=1;i<=q;i++)
	{
		qu[i].l=read(),qu[i].r=read();
		g[qu[i].l-1].push_back(qu[i].r);
	}

	for(int i=1;i<=n;i++)
		f[i]=Inf;
	f[0]=0;
	build(1,0,n);
	for(int i=0;i<n;i++)//pos
	{
		for(int j=0;j<g[i].size();j++)
		{
			chkmin(f[g[i][j]],query(1,0,n,i,g[i][j]-1)+sum[g[i][j]]);
			modify(1,0,n,g[i][j]);
		}
		chkmin(f[i+1],f[i]+(b[i+1]^1));
		modify(1,0,n,i+1);
	}

	printf("%d\n",f[n]);
	return 0;
}