DZY loves colors, and he enjoys painting.

On a colorful day, DZY gets a colorful ribbon, which consists of n units (they are numbered from 1 to n from left to right). The color of the i-th unit of the ribbon is i at first. It is colorful enough, but we still consider that the colorfulness of each unit is 0 at first.

DZY loves painting, we know. He takes up a paintbrush with color x and uses it to draw a line on the ribbon. In such a case some contiguous units are painted. Imagine that the color of unit i currently is y. When it is painted by this paintbrush, the color of the unit becomes x, and the colorfulness of the unit increases by |x - y|.

DZY wants to perform m operations, each operation can be one of the following:

  1. Paint all the units with numbers between l and r (both inclusive) with color x. 
  2. Ask the sum of colorfulness of the units between l and r (both inclusive). 



Can you help DZY?

Input

The first line contains two space-separated integers n, m (1 ≤ n, m ≤ 105).

Each of the next m lines begins with a integer type (1 ≤ type ≤ 2), which represents the type of this operation.

If type = 1, there will be 3 more integers l, r, x (1 ≤ l ≤ r ≤ n; 1 ≤ x ≤ 108) in this line, describing an operation 1.

If type = 2, there will be 2 more integers l, r (1 ≤ l ≤ r ≤ n) in this line, describing an operation 2.

Output

For each operation 2, print a line containing the answer — sum of colorfulness.

Examples

Input

3 3
1 1 2 4
1 2 3 5
2 1 3


Output

8


Input

3 4
1 1 3 4
2 1 1
2 2 2
2 3 3


Output

3
2
1


Input

10 6
1 1 5 3
1 2 7 9
1 10 10 11
1 3 8 12
1 1 10 3
2 1 10


Output

129

Note

In the first sample, the color of each unit is initially [1, 2, 3], and the colorfulness is [0, 0, 0].

After the first operation, colors become [4, 4, 3], colorfulness become [3, 2, 0].

After the second operation, colors become [4, 5, 5], colorfulness become [3, 3, 2].

So the answer to the only operation of type 2 is 8.
#include <bits/stdc++.h>
using namespace std;
const int maxn = (int)1e5 + 100;
const int maxbl = 320;
const int mod = (int)1e9 + 9;
const int maxlog = (int)20;
const int P = mod;
int n, m, color[4 * maxn];
long long t[4 * maxn], lz[4 * maxn];
void build(int v = 1, int tl = 1, int tr = n) {
  if (tl == tr) {
    color[v] = tl;
    return;
  }
  int tm = (tl + tr) >> 1;
  build(v + v, tl, tm);
  build(v + v + 1, tm + 1, tr);
}
void push(int v, int tl, int tr) {
  if (lz[v]) {
    t[v] += (tr - tl + 1) * lz[v];
    if (tl != tr) lz[v * 2] += lz[v], lz[v * 2 + 1] += lz[v];
    lz[v] = 0;
  }
  if (color[v] && tl != tr) color[v * 2] = color[v * 2 + 1] = color[v];
}
inline void upd(int l, int r, int x, int v = 1, int tl = 1, int tr = n) {
  push(v, tl, tr);
  if (l > r || tr < l || tl > r) return;
  if (l <= tl && tr <= r && color[v]) {
    lz[v] += abs(color[v] - x);
    color[v] = x;
    push(v, tl, tr);
    return;
  }
  int tm = (tl + tr) >> 1;
  upd(l, r, x, v + v, tl, tm);
  upd(l, r, x, v + v + 1, tm + 1, tr);
  t[v] = t[v + v] + t[v + v + 1];
  color[v] = 0;
  if (color[v + v] == color[v + v + 1]) color[v] = color[v + v];
}
long long get(int l, int r, int v = 1, int tl = 1, int tr = n) {
  push(v, tl, tr);
  if (tl > r || l > tr) return 0;
  if (tl >= l && tr <= r) return t[v];
  int tm = (tl + tr) / 2;
  return get(l, r, v * 2, tl, tm) + get(l, r, v * 2 + 1, tm + 1, tr);
}
int main() {
  cin >> n >> m;
  build();
  for (int i = 1; i <= m; i++) {
    int type, l, r, x;
    cin >> type;
    if (type == 1) {
      cin >> l >> r >> x;
      upd(l, r, x);
    } else {
      cin >> l >> r;
      cout << get(l, r) << "\n";
    }
  }
}
