#include <iostream>
#include <vector>
#include <algorithm>
#include <limits>
using namespace std;

template<typename Type>
class RMQ {
private:
	int size_; std::vector<Type> dat;
public:
	RMQ() : size_(0), dat(std::vector<Type>()) {};
	RMQ(int size__) {
		for (size_ = 1; size_ < size__; ) size_ <<= 1;
		dat.resize(size_ << 1, std::numeric_limits<Type>::max());
	}
	template<class T>
	RMQ(T begin_, T end_) {
		int n = end_ - begin_;
		for (size_ = 1; size_ < n; size_ <<= 1); dat.resize(size_ << 1, std::numeric_limits<Type>::max());
		for (int i = 0; i < n; i++) dat[i + size_] = *(begin_ + i);
		for (int i = size_ - 1; i > 0; i--) dat[i] = std::min(dat[i << 1], dat[(i << 1) + 1]);
	}
	inline unsigned size() { return size_; }
	inline void update(int i, Type x) {
		i += size_; dat[i] = x;
		while (i > 1) {
			i >>= 1;
			dat[i] = std::min(dat[i << 1], dat[i << 1 | 1]);
		}
	}
	inline Type query(int l, int r) {
		Type ret = std::numeric_limits<Type>::max();
		l += size_; r += size_;
		while (l != r) {
			if (l & 1) ret = std::min(ret, dat[l++]); l >>= 1;
			if (r & 1) ret = std::min(ret, dat[--r]); r >>= 1;
		}
		return ret;
	}
};

int n, q, b[200009], l[200009], r[200009], dp[200009], sum; vector<int>V[200009];

int main() {
	cin >> n; for (int i = 1; i <= n; i++) { cin >> b[i]; sum += b[i]; if (b[i] == 0) b[i] = -1; else b[i] = 1; }
	for (int i = 1; i <= n; i++) { b[i] += b[i - 1]; }
	cin >> q;
	for (int i = 1; i <= q; i++) {
		cin >> l[i] >> r[i];
		V[l[i]].push_back(r[i]);
	}
	RMQ<int>X(n + 3);
	for (int i = 1; i <= n; i++) { X.update(i, (1 << 30)); dp[i] = (1 << 30); }
	X.update(0, 0);
	int V1 = 0;
	for (int i = 1; i <= n; i++) {
		for (int j = 0; j < V[i].size(); j++) {
			int p1 = i, p2 = V[i][j];
			dp[p2] = min(dp[p2], X.query(p1, p2 + 1) - b[p2]);
			dp[p2] = min(dp[p2], V1 + (b[p1 - 1] - b[p2]));
			X.update(p2, dp[p2] + b[p2]);
		}
		V1 = min(V1, dp[i]);
	}
	int minx = 0; for (int i = 1; i <= n; i++) minx = min(minx, dp[i]);
	cout << sum + minx << endl;
	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 <iostream>
#include <vector>
#include <algorithm>
#include <limits>
using namespace std;

template<typename Type>
class RMQ {
private:
	int size_; std::vector<Type> dat;
public:
	RMQ() : size_(0), dat(std::vector<Type>()) {};
	RMQ(int size__) {
		for (size_ = 1; size_ < size__; ) size_ <<= 1;
		dat.resize(size_ << 1, std::numeric_limits<Type>::max());
	}
	template<class T>
	RMQ(T begin_, T end_) {
		int n = end_ - begin_;
		for (size_ = 1; size_ < n; size_ <<= 1); dat.resize(size_ << 1, std::numeric_limits<Type>::max());
		for (int i = 0; i < n; i++) dat[i + size_] = *(begin_ + i);
		for (int i = size_ - 1; i > 0; i--) dat[i] = std::min(dat[i << 1], dat[(i << 1) + 1]);
	}
	inline unsigned size() { return size_; }
	inline void update(int i, Type x) {
		i += size_; dat[i] = x;
		while (i > 1) {
			i >>= 1;
			dat[i] = std::min(dat[i << 1], dat[i << 1 | 1]);
		}
	}
	inline Type query(int l, int r) {
		Type ret = std::numeric_limits<Type>::max();
		l += size_; r += size_;
		while (l != r) {
			if (l & 1) ret = std::min(ret, dat[l++]); l >>= 1;
			if (r & 1) ret = std::min(ret, dat[--r]); r >>= 1;
		}
		return ret;
	}
};

int n, q, b[200009], l[200009], r[200009], dp[200009], sum; vector<int>V[200009];

int main() {
	cin >> n; for (int i = 1; i <= n; i++) { cin >> b[i]; sum += b[i]; if (b[i] == 0) b[i] = -1; else b[i] = 1; }
	for (int i = 1; i <= n; i++) { b[i] += b[i - 1]; }
	cin >> q;
	for (int i = 1; i <= q; i++) {
		cin >> l[i] >> r[i];
		V[l[i]].push_back(r[i]);
	}
	RMQ<int>X(n + 3);
	for (int i = 1; i <= n; i++) { X.update(i, (1 << 30)); dp[i] = (1 << 30); }
	X.update(0, 0);
	int V1 = 0;
	for (int i = 1; i <= n; i++) {
		for (int j = 0; j < V[i].size(); j++) {
			int p1 = i, p2 = V[i][j];
			dp[p2] = min(dp[p2], X.query(p1, p2 + 1) - b[p2]);
			dp[p2] = min(dp[p2], V1 + (b[p1 - 1] - b[p2]));
			X.update(p2, dp[p2] + b[p2]);
		}
		V1 = min(V1, dp[i]);
	}
	int minx = 0; for (int i = 1; i <= n; i++) minx = min(minx, dp[i]);
	cout << sum + minx << endl;
	return 0;
}
```