#include <vector>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
const int inf = 1012345678;
const int max_size = 262144;
int N, Q, l, r, a[200009], s[200009], dp[200009], seg[524296]; vector<int> g[200009];
void release(int pos, int val) {
	pos += max_size;
	seg[pos] = max(seg[pos], val);
	while (pos > 1) {
		pos >>= 1;
		seg[pos] = max(seg[pos * 2], seg[pos * 2 + 1]);
	}
}
int query(int a, int b, int k = 1, int l = 0, int r = max_size) {
	if (a <= l && r <= b) return seg[k];
	if (r <= a || b <= l) return -inf;
	int lc = query(a, b, k * 2, l, (l + r) >> 1);
	int rc = query(a, b, k * 2 + 1, (l + r) >> 1, r);
	return max(lc, rc);
}
int main() {
	cin.tie(0);
	ios_base::sync_with_stdio(false);
	cin >> N;
	for (int i = 0; i < N; i++) cin >> a[i], s[i + 1] = s[i] + a[i];
	cin >> Q;
	for (int i = 0; i < Q; i++) {
		cin >> l >> r; l--;
		g[l].push_back(r);
	}
	fill(seg, seg + max_size * 2, -inf);
	release(0, 0);
	for (int i = 0; i < N; i++) {
		sort(g[i].begin(), g[i].end(), greater<int>());
		for (int j : g[i]) release(j, query(i, j));
		release(i + 1, query(i, i + 1) + s[i] - s[i + 1] - a[i] + 1);
	}
	cout << N - query(N, N + 1) - s[N] << endl;
	return 0;
}
### Prompt

Please create a solution in Cpp to 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 <vector>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
const int inf = 1012345678;
const int max_size = 262144;
int N, Q, l, r, a[200009], s[200009], dp[200009], seg[524296]; vector<int> g[200009];
void release(int pos, int val) {
	pos += max_size;
	seg[pos] = max(seg[pos], val);
	while (pos > 1) {
		pos >>= 1;
		seg[pos] = max(seg[pos * 2], seg[pos * 2 + 1]);
	}
}
int query(int a, int b, int k = 1, int l = 0, int r = max_size) {
	if (a <= l && r <= b) return seg[k];
	if (r <= a || b <= l) return -inf;
	int lc = query(a, b, k * 2, l, (l + r) >> 1);
	int rc = query(a, b, k * 2 + 1, (l + r) >> 1, r);
	return max(lc, rc);
}
int main() {
	cin.tie(0);
	ios_base::sync_with_stdio(false);
	cin >> N;
	for (int i = 0; i < N; i++) cin >> a[i], s[i + 1] = s[i] + a[i];
	cin >> Q;
	for (int i = 0; i < Q; i++) {
		cin >> l >> r; l--;
		g[l].push_back(r);
	}
	fill(seg, seg + max_size * 2, -inf);
	release(0, 0);
	for (int i = 0; i < N; i++) {
		sort(g[i].begin(), g[i].end(), greater<int>());
		for (int j : g[i]) release(j, query(i, j));
		release(i + 1, query(i, i + 1) + s[i] - s[i + 1] - a[i] + 1);
	}
	cout << N - query(N, N + 1) - s[N] << endl;
	return 0;
}
```