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 = 100005;
long long myabs(long long x) {
  if (x < 0) return -x;
  return x;
}
struct SEG {
  int l, r, mid;
  long long c, sum, add;
} seg[MAXN << 2];
void build(int rt, int l, int r) {
  seg[rt].l = l, seg[rt].r = r;
  seg[rt].mid = (l + r) >> 1;
  seg[rt].c = -1;
  seg[rt].add = seg[rt].sum = 0;
  if (l == r) {
    seg[rt].c = l;
    return;
  }
  build(rt << 1, l, seg[rt].mid);
  build(rt << 1 | 1, seg[rt].mid + 1, r);
}
void pushdown(int rt) {
  if (seg[rt].add) {
    seg[rt << 1].add += seg[rt].add;
    seg[rt << 1 | 1].add += seg[rt].add;
    seg[rt << 1].sum += seg[rt].add * (seg[rt << 1].r - seg[rt << 1].l + 1);
    seg[rt << 1 | 1].sum +=
        seg[rt].add * (seg[rt << 1 | 1].r - seg[rt << 1 | 1].l + 1);
    seg[rt << 1].c = seg[rt << 1 | 1].c = seg[rt].c;
    seg[rt].add = 0;
  }
}
void pushup(int rt) {
  seg[rt].sum = seg[rt << 1].sum + seg[rt << 1 | 1].sum;
  if (seg[rt << 1].c == seg[rt << 1 | 1].c)
    seg[rt].c = seg[rt << 1].c;
  else
    seg[rt].c = -1;
}
void update(int rt, int l, int r, long long c) {
  if (l <= seg[rt].l && seg[rt].r <= r && seg[rt].c >= 0) {
    seg[rt].add += myabs(seg[rt].c - c);
    seg[rt].sum += myabs(seg[rt].c - c) * (seg[rt].r - seg[rt].l + 1);
    seg[rt].c = c;
    return;
  }
  pushdown(rt);
  if (l <= seg[rt].mid) update(rt << 1, l, r, c);
  if (seg[rt].mid < r) update(rt << 1 | 1, l, r, c);
  pushup(rt);
}
long long query(int rt, int l, int r) {
  if (l <= seg[rt].l && seg[rt].r <= r) return seg[rt].sum;
  pushdown(rt);
  long long res = 0;
  if (l <= seg[rt].mid) res += query(rt << 1, l, r);
  if (seg[rt].mid < r) res += query(rt << 1 | 1, l, r);
  pushup(rt);
  return res;
}
int n, m;
int main() {
  while (scanf("%d%d", &n, &m) != EOF) {
    build(1, 1, n);
    int c, l, r;
    long long x;
    for (int i = 1; i <= m; i++) {
      scanf("%d%d%d", &c, &l, &r);
      if (c == 2)
        printf("%I64d\n", query(1, l, r));
      else {
        scanf("%I64d", &x);
        update(1, l, r, x);
      }
    }
  }
  return 0;
}
