Nagini, being a horcrux You-know-who created with the murder of Bertha Jorkins, has accumulated its army of snakes and is launching an attack on Hogwarts school. 

Hogwarts' entrance can be imagined as a straight line (x-axis) from 1 to 105. Nagini is launching various snakes at the Hogwarts entrance. Each snake lands parallel to the entrance, covering a segment at a distance k from x = l to x = r. Formally, each snake can be imagined as being a line segment between points (l, k) and (r, k). Note that k can be both positive and negative, but not 0.

Let, at some x-coordinate x = i, there be snakes at point (i, y1) and point (i, y2), such that y1 > 0 and y2 < 0. Then, if for any point (i, y3) containing a snake such that y3 > 0, y1 ≤ y3 holds and for any point (i, y4) containing a snake such that y4 < 0, |y2| ≤ |y4| holds, then the danger value at coordinate x = i is y1 + |y2|. If no such y1 and y2 exist, danger value is 0. 

Harry wants to calculate the danger value of various segments of the Hogwarts entrance. Danger value for a segment [l, r) of the entrance can be calculated by taking the sum of danger values for each integer x-coordinate present in the segment.

Formally, you have to implement two types of queries:

  * 1 l r k: a snake is added parallel to entrance from x = l to x = r at y-coordinate y = k (l inclusive, r exclusive). 
  * 2 l r: you have to calculate the danger value of segment l to r (l inclusive, r exclusive). 

Input

First line of input contains a single integer q (1 ≤ q ≤ 5·104) denoting the number of queries.

Next q lines each describe a query. Each query description first contains the query type typei (1 ≤ typei ≤ 2). This is followed by further description of the query. In case of the type being 1, it is followed by integers li, ri and ki (<image>,  - 109 ≤ ki ≤ 109, k ≠ 0). Otherwise, it just contains two integers, li and ri (1 ≤ li < ri ≤ 105).

Output

Output the answer for each query of type 2 in a separate line.

Examples

Input

3
1 1 10 10
1 2 4 -7
2 1 10


Output

34


Input

7
1 2 3 5
1 1 10 10
1 4 5 -5
2 4 8
1 1 10 -10
2 4 8
2 1 10


Output

15
75
170

Note

In the first sample case, the danger value for x-coordinates 1 is 0 as there is no y2 satisfying the above condition for x = 1.

Danger values for x-coordinates 2 and 3 is 10 + | - 7| = 17.

Danger values for x-coordinates 4 to 9 is again 0 as there is no y2 satisfying the above condition for these coordinates.

Thus, total danger value is 17 + 17 = 34.
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast,no-stack-protector")
#pragma GCC target("avx,tune=native")
#pragma GCC diagnostic ignored "-Wunused-result"
using namespace std;
template <class T>
inline bool chkmax(T &a, const T &b) {
  return b > a ? a = b, true : false;
}
template <class T>
inline bool chkmin(T &a, const T &b) {
  return b < a ? a = b, true : false;
}
const int N = 1e5 + 10;
const int INF = 1e9 + 10;
int au[N], ad[N];
template <class T>
void _R(T &x) {
  cin >> x;
}
void _R(int &x) { scanf("%d", &x); }
void _R(int64_t &x) { scanf("%" PRId64, &x); }
void _R(double &x) { scanf("%lf", &x); }
void _R(char &x) { scanf(" %c", &x); }
void _R(char *x) { scanf("%s", x); }
void R() {}
template <class T, class... U>
void R(T &head, U &...tail) {
  _R(head);
  R(tail...);
}
template <class T>
void _W(const T &x) {
  cout << x;
}
void _W(const int &x) { printf("%d", x); }
void _W(const int64_t &x) { printf("%" PRId64, x); }
void _W(const double &x) { printf("%.16f\n", x); }
void _W(const char &x) { putchar(x); }
void _W(const char *x) { printf("%s", x); }
template <class T>
void _W(const vector<T> &x) {
  for (auto i = x.begin(); i != x.end(); _W(*i++)) {
    if (i != x.cbegin()) putchar(' ');
  }
}
void W() {}
template <class T, class... U>
void W(const T &head, const U &...tail) {
  _W(head);
  putchar(sizeof...(tail) ? ' ' : '\n');
  W(tail...);
}
int main() {
  for (int i = 1; i < N; i++) au[i] = ad[i] = INF;
  int q;
  R(q);
  while (q--) {
    int op, l, r;
    R(op, l, r);
    r--;
    if (op == 1) {
      int y;
      R(y);
      if (y > 0) {
        for (int i = l; i <= r; i++) chkmin(au[i], y);
      } else {
        y = -y;
        for (int i = l; i <= r; i++) chkmin(ad[i], y);
      }
    } else {
      long long int ans = 0;
      for (int i = l; i <= r; i++)
        if (au[i] < INF && ad[i] < INF) ans += au[i] + ad[i];
      W(ans);
    }
  }
  return 0;
}
