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 <bits/stdc++.h>
// #define INF 5000000000000000000
#define ll long long
#define pll pair<ll, ll>
using namespace std;

const ll MAX_N = 1 << 20;
const ll INF = pow(2, 31) - 1;
ll n;
vector<ll> dat(MAX_N, 0);
vector<ll> delay(MAX_N, 0);

void init(ll _n) {
  ll temp = 1;
  while (_n >= temp) {
    temp *= 2;
  }
  n = temp;
}

void evaluate(ll k) {
  dat.at(k) += delay.at(k);
  if (delay.at(k) == 0 || k >= n - 1) {
    delay.at(k) = 0;
    return;
  }
  delay.at(k * 2 + 1) += delay.at(k);
  delay.at(k * 2 + 2) += delay.at(k);
  delay.at(k) = 0;
}

void update(ll x, ll y, ll s, ll k = 0, ll l = 0, ll r = n) {
  evaluate(k);
  if (r <= x || y <= l) {
    return;
  }
  if (x <= l && r <= y) {
    delay.at(k) += s;
    evaluate(k);
  }
  else {
    update(x, y, s, k * 2 + 1, l, (l + r) / 2);
    update(x, y, s, k * 2 + 2, (l + r) / 2, r);
    dat.at(k) = min(dat.at(k * 2 + 1), dat.at(k * 2 + 2));
  }
}

ll query(ll x, ll y, ll k = 0, ll l = 0, ll r = n) {
  evaluate(k);
  if (r <= x || y <= l) {
    return INF;
  }
  if (x <= l && r <= y) {
    return dat.at(k);
  }
  else {
    ll lm = query(x, y, k * 2 + 1, l, (l + r) / 2);
    ll rm = query(x, y, k * 2 + 2, (l + r) / 2, r);
    return min(lm, rm);
  }
}

int main() {
  ll q;
  cin >> n >> q;
  init(n);
  for (ll i = 0; i < q; ++i) {
    ll com , x, y;
    cin >> com >> x >> y;
    if (com == 0) {
      ll s;
      cin >> s;
      update(x, y + 1, s);
    }
    else {
      cout << query(x, y + 1) << "\n";;
    }
    // for (ll i = 0; i < n * 2 - 1; ++i) {
    //   cout << dat.at(i) << " ";
    // }
    // cout << "\n";
  }
}