Write a program which manipulates a sequence $A$ = {$a_0, a_1, ..., a_{n-1}$} with the following operations:

* $add(s, t, x)$ : add $x$ to $a_s, a_{s+1}, ..., a_t$.
* $find(s, t)$ : report the minimum value in $a_s, a_{s+1}, ..., a_t$.



Note that the initial values of $a_i ( i = 0, 1, ..., n-1 )$ are 0.

Constraints

* $1 ≤ n ≤ 100000$
* $1 ≤ q ≤ 100000$
* $0 ≤ s ≤ t < n$
* $-1000 ≤ x ≤ 1000$

Input


$n$ $q$
$query_1$
$query_2$
:
$query_q$


In the first line, $n$ (the number of elements in $A$) and $q$ (the number of queries) are given. Then, $i$th query $query_i$ is given in the following format:


0 $s$ $t$ $x$


or


1 $s$ $t$


The first digit represents the type of the query. '0' denotes $add(s, t, x)$ and '1' denotes $find(s, t)$.

Output

For each $find$ query, print the minimum value.

Example

Input

6 7
0 1 3 1
0 2 4 -2
1 0 5
1 0 1
0 3 5 3
1 3 4
1 0 5


Output

-2
0
1
-1
#include <bits/stdc++.h>
using namespace std;

const int MAXn = 200005, MAX = 2147483647;
int n, q, t1, t2, t3, t4, h, t[MAXn], d[MAXn];
bool tag[MAXn];

int msb(int val) {
	int i = 0;
	while (val) val >>= 1, i++;
	return i;
}

void apply(int p, int val) {
	t[p] += val;
	if (p < n) d[p] += val, tag[p] = 1;
}

void push(int p) {
	for (int s = h; s > 0; s--) {
		int i = p >> s;
		if (tag[i]) {
			apply(i << 1, d[i]);
			apply(i << 1 | 1, d[i]);
			tag[i] = 0, d[i] = 0;
		}
	}
}

void build(int p) {
	while (p > 1) p >>= 1, t[p] = min(t[p << 1], t[p << 1 | 1]) + d[p];
}

void update(int l, int r, int val) {
	l += n, r += n;
	int l0 = l, r0 = r;
	for (; l < r; l >>= 1, r >>= 1) {
		if (l & 1) apply(l++, val);
		if (r & 1) apply(--r, val);
	}
	build(l0);
	build(r0 - 1);
}

int query(int l, int r) {
	l += n, r += n;
	push(l);
	push(r - 1);
	int ans = MAX;
	for (; l < r; l >>= 1, r >>= 1) {
		if (l & 1) ans = min(ans, t[l++]);
		if (r & 1) ans = min(ans, t[--r]);
	}
	return ans;
}

int main() {
	cin >> n >> q;
	h = msb(n);
	for (int i = 0; i < q; i++) {
		cin >> t1;
		if (t1 == 0) {
			cin >> t2 >> t3 >> t4;
			update(t2, t3 + 1, t4);
//			for (int j = 1; j < 2 * n; j++) cout << t[j] << ' ';
//			cout << endl;
//			for (int j = 1; j < 2 * n; j++) cout << d[j] << ' ';
//			cout << endl;
		}
		else {
			cin >> t2 >> t3;
			cout << query(t2, t3 + 1) << endl;
//			for (int j = 1; j < 2 * n; j++) cout << t[j] << ' ';
//			cout << endl;
//			for (int j = 1; j < 2 * n; j++) cout << d[j] << ' ';
//			cout << endl;
		}
	}
}
