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;
long long sum[101111 << 2], col[101111 << 2], add[101111 << 2], n, m, a, b, c,
    t;
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;
}
void pushdown(long long l, long long r, long long rt) {
  if (add[rt]) {
    long long mid = (l + r) >> 1;
    sum[rt << 1] += (mid - l + 1) * add[rt];
    sum[rt << 1 | 1] += (r - mid) * add[rt];
    add[rt << 1] += add[rt];
    add[rt << 1 | 1] += add[rt];
    if (col[rt]) col[rt << 1] = col[rt << 1 | 1] = col[rt];
    add[rt] = 0;
  }
}
void pushup(long long rt) {
  if (col[rt << 1] == col[rt << 1 | 1])
    col[rt] = col[rt << 1];
  else
    col[rt] = 0;
  sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
}
void build(long long l, long long r, long long rt) {
  if (l == r) {
    col[rt] = l;
    return;
  }
  long long mid = (l + r) >> 1;
  build(l, mid, rt << 1);
  build(mid + 1, r, rt << 1 | 1);
}
long long query(long long l, long long r, long long rt) {
  if (a <= l && r <= b) return sum[rt];
  long long ans = 0;
  long long mid = (l + r) >> 1;
  pushdown(l, r, rt);
  if (a <= mid) ans += query(l, mid, rt << 1);
  if (mid < b) ans += query(mid + 1, r, rt << 1 | 1);
  return ans;
}
void update(long long l, long long r, long long rt) {
  if (a <= l && r <= b && col[rt]) {
    sum[rt] += (r - l + 1) * abs(col[rt] - c);
    add[rt] += abs(col[rt] - c);
    col[rt] = c;
    return;
  }
  long long mid = (l + r) >> 1;
  pushdown(l, r, rt);
  if (a <= mid) update(l, mid, rt << 1);
  if (mid < b) update(mid + 1, r, rt << 1 | 1);
  pushup(rt);
}
int main() {
  n = read(), m = read();
  build(1, n, 1);
  while (m--) {
    t = read();
    if (--t) {
      a = read(), b = read();
      printf("%I64d\n", query(1, n, 1));
    } else {
      a = read(), b = read(), c = read();
      update(1, n, 1);
    }
  }
}
