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 INF = 0x3f3f3f3f, MAXN = 100000 + 10;
int col[MAXN << 1];
long long sum[MAXN << 1], add[MAXN << 1];
int id(int x, int y) { return x + y | x != y; }
void bulid(int x, int y) {
  if (x == y) {
    int u = id(x, y);
    col[u] = x;
    return;
  }
  int m = (x + y) >> 1;
  bulid(x, m);
  bulid(m + 1, y);
}
void push_up(int x, int y, int m) {
  int u = id(x, y), l = id(x, m), r = id(m + 1, y);
  sum[u] = sum[l] + sum[r];
  if (col[l] == col[r])
    col[u] = col[l];
  else
    col[u] = -1;
}
void down(int x, int y, long long v) {
  int u = id(x, y);
  add[u] += v;
  sum[u] += (long long)v * (y - x + 1);
}
void push_down(int x, int y, int m) {
  int u = id(x, y);
  if (~col[u]) col[id(x, m)] = col[id(m + 1, y)] = col[u];
  if (add[u]) down(x, m, add[u]), down(m + 1, y, add[u]), add[u] = 0;
}
void update(int x, int y, int ql, int qr, int v) {
  if (ql <= x && y <= qr) {
    if (~col[id(x, y)]) {
      int u = id(x, y);
      add[u] += abs(v - col[u]);
      sum[u] += (long long)abs(v - col[u]) * (y - x + 1);
      col[u] = v;
      return;
    }
  }
  int m = (x + y) >> 1;
  push_down(x, y, m);
  if (ql <= m) update(x, m, ql, qr, v);
  if (qr > m) update(m + 1, y, ql, qr, v);
  push_up(x, y, m);
}
long long query(int x, int y, int ql, int qr) {
  if (ql <= x && y <= qr) return sum[id(x, y)];
  int m = (x + y) >> 1;
  push_down(x, y, m);
  long long ans = 0;
  if (ql <= m) ans += query(x, m, ql, qr);
  if (qr > m) ans += query(m + 1, y, ql, qr);
  return ans;
}
int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  memset(col, -1, sizeof(col));
  bulid(1, n);
  while (m--) {
    int op, x, y;
    scanf("%d%d%d", &op, &x, &y);
    if (1 == op) {
      int v;
      scanf("%d", &v);
      update(1, n, x, y, v);
    } else
      printf("%I64d\n", query(1, n, x, y));
  }
  return 0;
}
