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;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, m;
struct TREE {
  int l, r, col;
  long long sum, add;
} rt[100100 << 2];
void build(int x, int l, int r) {
  rt[x].l = l;
  rt[x].r = r;
  rt[x].sum = rt[x].add = rt[x].col = 0;
  if (l == r) {
    rt[x].col = l;
    return;
  }
  int mid = (l + r) >> 1;
  build((x << 1), l, mid);
  build((x << 1 | 1), mid + 1, r);
}
void pushup(int x) {
  rt[x].sum = rt[(x << 1)].sum + rt[(x << 1 | 1)].sum;
  if (rt[(x << 1)].col == rt[(x << 1 | 1)].col)
    rt[x].col = rt[(x << 1)].col;
  else
    rt[x].col = 0;
}
void pushdown(int x) {
  if (rt[x].add) {
    rt[(x << 1)].add += rt[x].add;
    rt[(x << 1)].sum += (rt[(x << 1)].r - rt[(x << 1)].l + 1) * rt[x].add;
    rt[(x << 1 | 1)].add += rt[x].add;
    rt[(x << 1 | 1)].sum +=
        (rt[(x << 1 | 1)].r - rt[(x << 1 | 1)].l + 1) * rt[x].add;
    rt[x].add = 0;
    if (rt[x].col) rt[(x << 1)].col = rt[(x << 1 | 1)].col = rt[x].col;
  }
}
void update(int x, int l, int r, int v) {
  if (rt[x].l >= l && rt[x].r <= r && rt[x].col) {
    rt[x].sum += (long long)(rt[x].r - rt[x].l + 1) * abs(v - rt[x].col);
    rt[x].add += abs(v - rt[x].col);
    rt[x].col = v;
    return;
  }
  pushdown(x);
  int mid = (rt[x].l + rt[x].r) >> 1;
  if (l <= mid) update((x << 1), l, r, v);
  if (r > mid) update((x << 1 | 1), l, r, v);
  pushup(x);
}
long long query(int x, int l, int r) {
  if (rt[x].l >= l && rt[x].r <= r) return rt[x].sum;
  pushdown(x);
  int mid = (rt[x].l + rt[x].r) >> 1;
  long long ans = 0;
  if (l <= mid) ans += query((x << 1), l, r);
  if (r > mid) ans += query((x << 1 | 1), l, r);
  return ans;
}
int main() {
  n = read();
  m = read();
  build(1, 1, n);
  int p, l, r, c;
  for (int i = 1; i <= m; i++) {
    p = read();
    if (p == 1) {
      l = read();
      r = read();
      c = read();
      update(1, l, r, c);
    } else {
      l = read();
      r = read();
      printf("%lld\n", query(1, l, r));
    }
  }
  return 0;
}
