#include <iostream>
#include <vector>
#define inf 1000000000

using namespace std;

struct SegTree{
	int size;
	vector<int> seg;
	
	SegTree(){}
	SegTree(int size){
		this->size = size;
		seg.resize(1<<(size+1));
	}
	
	void init()
	{
		for(int i = 0; i < (1<<(size+1)); i++) seg[i] = inf;
	}
	
	void update(int i, int val)
	{
		i += (1 << size);
		seg[i] = val;
		while(i > 1){
			i /= 2;
			seg[i] = min(seg[i*2], seg[i*2+1]);
		}
	}

	int query(int a, int b, int k, int l, int r)
	{
		if(b < l || r < a) return inf;
		if(a <= l && r <= b) return seg[k];
		int lval = query(a, b, k*2, l, (l+r)/2);
		int rval = query(a, b, k*2+1, (l+r)/2+1, r);
		return min(lval, rval);
	}
	int query(int a, int b)
	{
		return query(a, b, 1, 0, (1<<size)-1);
	}
};

int N, Q;
int b[200005];

vector<int> vec[200005];
SegTree seg(18);

int main(void)
{
	cin >> N;
	for(int i = 1; i <= N; i++) cin >> b[i];
	
	cin >> Q;
	int l, r;
	for(int i = 0; i < Q; i++){
		cin >> l >> r;
		vec[l].push_back(r);
	}
	
	seg.init();
	seg.update(0, 0);
	
	for(int i = 1; i <= N; i++){
		for(int j = 0; j < vec[i].size(); j++){
			seg.update(vec[i][j], min(seg.query(vec[i][j], vec[i][j]), seg.query(i-1, vec[i][j]-1)));
		}
		if(b[i]) seg.update(i, min(seg.query(i, i), seg.query(i-1, i-1) + 1));
		else seg.update(i, min(seg.query(i, i), seg.query(i-1, i-1) - 1));
	}
	
	int ans = seg.query(N, N);
	for(int i = 1; i <= N; i++) if(b[i] == 0) ans++;
	cout << ans << endl;
	
	return 0;
}
### Prompt

In Cpp, your task is to solve the following 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

### Response

```cpp
#include <iostream>
#include <vector>
#define inf 1000000000

using namespace std;

struct SegTree{
	int size;
	vector<int> seg;
	
	SegTree(){}
	SegTree(int size){
		this->size = size;
		seg.resize(1<<(size+1));
	}
	
	void init()
	{
		for(int i = 0; i < (1<<(size+1)); i++) seg[i] = inf;
	}
	
	void update(int i, int val)
	{
		i += (1 << size);
		seg[i] = val;
		while(i > 1){
			i /= 2;
			seg[i] = min(seg[i*2], seg[i*2+1]);
		}
	}

	int query(int a, int b, int k, int l, int r)
	{
		if(b < l || r < a) return inf;
		if(a <= l && r <= b) return seg[k];
		int lval = query(a, b, k*2, l, (l+r)/2);
		int rval = query(a, b, k*2+1, (l+r)/2+1, r);
		return min(lval, rval);
	}
	int query(int a, int b)
	{
		return query(a, b, 1, 0, (1<<size)-1);
	}
};

int N, Q;
int b[200005];

vector<int> vec[200005];
SegTree seg(18);

int main(void)
{
	cin >> N;
	for(int i = 1; i <= N; i++) cin >> b[i];
	
	cin >> Q;
	int l, r;
	for(int i = 0; i < Q; i++){
		cin >> l >> r;
		vec[l].push_back(r);
	}
	
	seg.init();
	seg.update(0, 0);
	
	for(int i = 1; i <= N; i++){
		for(int j = 0; j < vec[i].size(); j++){
			seg.update(vec[i][j], min(seg.query(vec[i][j], vec[i][j]), seg.query(i-1, vec[i][j]-1)));
		}
		if(b[i]) seg.update(i, min(seg.query(i, i), seg.query(i-1, i-1) + 1));
		else seg.update(i, min(seg.query(i, i), seg.query(i-1, i-1) - 1));
	}
	
	int ans = seg.query(N, N);
	for(int i = 1; i <= N; i++) if(b[i] == 0) ans++;
	cout << ans << endl;
	
	return 0;
}
```