Problem: 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

Solution (in CPP):

Solution:
#include <iostream>
#include <algorithm>
#include <vector>
#include <numeric> // accumulate
#include <queue>
#include <string>
#include <cmath> // sqrt
using namespace std;
typedef long long ll;
const int INF = 1e9;

struct LazySegmentTree {
private:
    int n;
    vector<ll> node, lazy;

public:
    LazySegmentTree(vector<ll> v) {
        int sz = (int)v.size();
        n = 1; while(n < sz) n *= 2;
        node.resize(2*n-1);
        lazy.resize(2*n-1, 0);

        for(int i=0; i<sz; i++) node[i+n-1] = v[i];
        for(int i=n-2; i>=0; i--) node[i] = min(node[i*2+1], node[i*2+2]);
    }

    void eval(int k, int l, int r) {
        if(lazy[k] != 0) {
            node[k] += lazy[k];
            if(r-l > 1) {
                lazy[2*k+1] += lazy[k];
                lazy[2*k+2] += lazy[k];
            }
            lazy[k] = 0;
        }
    }

    void add(int a, int b, ll x, int k=0, int l=0, int r=-1) {
        if(r < 0) r = n;
        eval(k, l, r);
        if(b <= l || r <= a) return;
        if(a <= l && r <= b) {
            lazy[k] += x;
            eval(k, l, r);
        }
        else {
            add(a, b, x, 2*k+1, l, (l+r)/2);
            add(a, b, x, 2*k+2, (l+r)/2, r);
            node[k] = min(node[2*k+1], node[2*k+2]);
        }
    }

    ll getmin(int a, int b, int k=0, int l=0, int r=-1) {
        if(r < 0) r = n;
        eval(k, l, r);
        if(b <= l || r <= a) return INF;
        if(a <= l && r <= b) return node[k];
        ll vl = getmin(a, b, 2*k+1, l, (l+r)/2);
        ll vr = getmin(a, b, 2*k+2, (l+r)/2, r);
        return min(vl, vr);
    }

};
 
int main() {
    int n, q;
    cin >> n >> q;
    LazySegmentTree seg(vector<ll> (n, 0));

    for (int i = 0; i < q; i++) {
        int type, left, right;
        cin >> type >> left >> right;
        if (type == 0) {
            int val;
            cin >> val;
            seg.add(left, right+1, val);
        }
        else {
            cout << seg.getmin(left, right+1) << endl;
        }

    }

}