#include <bits/stdc++.h>

using namespace std;

const int maxn = 200005;

struct node
{
	int tag,mi;
}T[maxn * 4];

vector<int> lk[maxn];
int b[maxn],n,q;

void mark(int jd,int tag)
{
	T[jd].tag += tag,T[jd].mi += tag;
}

void down(int jd)
{
	if (!T[jd].tag) return;
	mark(jd << 1,T[jd].tag),mark(jd << 1 | 1,T[jd].tag),T[jd].tag = 0;
}

int query(int l,int r,int i,int j,int jd)
{
	if (j < l || i > r) return (1 << 30);
	if (i <= l && r <= j) return T[jd].mi;
	int mid = l + r >> 1;
	down(jd);
	return min(query(l,mid,i,j,jd << 1),query(mid + 1,r,i,j,jd << 1 | 1));
}

void update(int l,int r,int p,int v,int jd)
{
	if (l == r) T[jd].mi = min(T[jd].mi,v); else
	{
		int mid = l + r >> 1;
		down(jd);
		if (p <= mid) update(l,mid,p,v,jd << 1); else update(mid + 1,r,p,v,jd << 1 | 1);
		T[jd].mi = min(T[jd << 1].mi,T[jd << 1 | 1].mi);
	}
}

void add(int l,int r,int i,int j,int ad,int jd)
{
	if (j < l || i > r) return;
	if (i <= l && r <= j) mark(jd,ad); else
	{
		int mid = l + r >> 1;
		down(jd);
		add(l,mid,i,j,ad,jd << 1),add(mid + 1,r,i,j,ad,jd << 1 | 1);
		T[jd].mi = min(T[jd << 1].mi,T[jd << 1 | 1].mi);
	}
}

void build(int l,int r,int jd)
{
	if (l == r)
	{
		if (!l) T[jd].mi = 0; else T[jd].mi = (1 << 30);
		return;
	}
	int mid = l + r >> 1;
	build(l,mid,jd << 1),build(mid + 1,r,jd << 1 | 1);
	T[jd].mi = min(T[jd << 1].mi,T[jd << 1 | 1].mi);
}

int main()
{
	scanf("%d", &n);
	int sum = 0;
	for(int i = 1;i <= n;i ++)
		scanf("%d", &b[i]),sum += (!b[i]);
	int q;
	scanf("%d", &q);
	for(int l,r;q;q --)
	{
		scanf("%d%d", &l, &r);
		lk[l].push_back(r);
	}
	build(0,n,1);
	for(int i = 1;i <= n;i ++)
	{
		for(int j = 0;j < lk[i].size();j ++)
		{
			int r = lk[i][j];
			update(0,n,r,query(0,n,0,r - 1,1),1);
		}
		add(0,n,0,i - 1,b[i] ? 1 : -1,1);
	}
	printf("%d\n", T[1].mi + sum);
	return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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

### Response

```cpp
#include <bits/stdc++.h>

using namespace std;

const int maxn = 200005;

struct node
{
	int tag,mi;
}T[maxn * 4];

vector<int> lk[maxn];
int b[maxn],n,q;

void mark(int jd,int tag)
{
	T[jd].tag += tag,T[jd].mi += tag;
}

void down(int jd)
{
	if (!T[jd].tag) return;
	mark(jd << 1,T[jd].tag),mark(jd << 1 | 1,T[jd].tag),T[jd].tag = 0;
}

int query(int l,int r,int i,int j,int jd)
{
	if (j < l || i > r) return (1 << 30);
	if (i <= l && r <= j) return T[jd].mi;
	int mid = l + r >> 1;
	down(jd);
	return min(query(l,mid,i,j,jd << 1),query(mid + 1,r,i,j,jd << 1 | 1));
}

void update(int l,int r,int p,int v,int jd)
{
	if (l == r) T[jd].mi = min(T[jd].mi,v); else
	{
		int mid = l + r >> 1;
		down(jd);
		if (p <= mid) update(l,mid,p,v,jd << 1); else update(mid + 1,r,p,v,jd << 1 | 1);
		T[jd].mi = min(T[jd << 1].mi,T[jd << 1 | 1].mi);
	}
}

void add(int l,int r,int i,int j,int ad,int jd)
{
	if (j < l || i > r) return;
	if (i <= l && r <= j) mark(jd,ad); else
	{
		int mid = l + r >> 1;
		down(jd);
		add(l,mid,i,j,ad,jd << 1),add(mid + 1,r,i,j,ad,jd << 1 | 1);
		T[jd].mi = min(T[jd << 1].mi,T[jd << 1 | 1].mi);
	}
}

void build(int l,int r,int jd)
{
	if (l == r)
	{
		if (!l) T[jd].mi = 0; else T[jd].mi = (1 << 30);
		return;
	}
	int mid = l + r >> 1;
	build(l,mid,jd << 1),build(mid + 1,r,jd << 1 | 1);
	T[jd].mi = min(T[jd << 1].mi,T[jd << 1 | 1].mi);
}

int main()
{
	scanf("%d", &n);
	int sum = 0;
	for(int i = 1;i <= n;i ++)
		scanf("%d", &b[i]),sum += (!b[i]);
	int q;
	scanf("%d", &q);
	for(int l,r;q;q --)
	{
		scanf("%d%d", &l, &r);
		lk[l].push_back(r);
	}
	build(0,n,1);
	for(int i = 1;i <= n;i ++)
	{
		for(int j = 0;j < lk[i].size();j ++)
		{
			int r = lk[i][j];
			update(0,n,r,query(0,n,0,r - 1,1),1);
		}
		add(0,n,0,i - 1,b[i] ? 1 : -1,1);
	}
	printf("%d\n", T[1].mi + sum);
	return 0;
}

```