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 long long read() {
  long long 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 * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, m;
struct data {
  int l, r, c, ctag, size;
  long long ans, atag;
} t[400005];
void pushdown(int k) {
  if (t[k].l == t[k].r) return;
  int ctag = t[k].ctag;
  long long atag = t[k].atag;
  t[k].ctag = t[k].atag = 0;
  if (ctag) {
    t[k << 1].c = ctag;
    t[k << 1 | 1].c = ctag;
    t[k << 1].ctag = ctag;
    t[k << 1 | 1].ctag = ctag;
  }
  if (atag) {
    t[k << 1].atag += atag;
    t[k << 1 | 1].atag += atag;
    t[k << 1].ans += t[k << 1].size * atag;
    t[k << 1 | 1].ans += t[k << 1 | 1].size * atag;
  }
}
inline void pushup(int k) {
  if (t[k << 1].c == t[k << 1 | 1].c)
    t[k].c = t[k << 1].c;
  else
    t[k].c = -1;
  t[k].ans = t[k << 1].ans + t[k << 1 | 1].ans;
}
long long query(int k, int x, int y) {
  pushdown(k);
  int l = t[k].l, r = t[k].r;
  if (l == x && y == r) return t[k].ans;
  int mid = (l + r) >> 1;
  if (y <= mid)
    return query(k << 1, x, y);
  else if (x > mid)
    return query(k << 1 | 1, x, y);
  else
    return query(k << 1, x, mid) + query(k << 1 | 1, mid + 1, y);
  pushup(k);
}
void update(int k, int x, int y, int val) {
  pushdown(k);
  int l = t[k].l, r = t[k].r;
  if (l == x && y == r && t[k].c != -1) {
    long long tmp = abs(t[k].c - val);
    t[k].ans += t[k].size * tmp;
    t[k].atag += tmp;
    t[k].c = val;
    t[k].ctag = val;
    return;
  }
  int mid = (l + r) >> 1;
  if (y <= mid)
    update(k << 1, x, y, val);
  else if (x > mid)
    update(k << 1 | 1, x, y, val);
  else {
    update(k << 1, x, mid, val);
    update(k << 1 | 1, mid + 1, y, val);
  }
  pushup(k);
}
void build(int k, int l, int r) {
  t[k].l = l;
  t[k].r = r;
  t[k].size = r - l + 1;
  t[k].c = -1;
  if (l == r) {
    t[k].c = l;
    return;
  }
  int mid = (l + r) >> 1;
  build(k << 1, l, mid);
  build(k << 1 | 1, mid + 1, r);
}
int main() {
  n = read();
  m = read();
  build(1, 1, n);
  for (int i = 1; i <= m; i++) {
    int f = read(), l = read(), r = read();
    if (f == 1) {
      int x = read();
      update(1, l, r, x);
    } else {
      printf("%I64d\n", query(1, l, r));
    }
  }
  return 0;
}
