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 N = 110000;
struct Node {
  int l, r, co;
  long long val, lz;
  Node() {
    lz = 0LL;
    co = -1;
    val = 0LL;
  }
} t[N << 2];
int n, m, l, r, x, flag;
long long ABS(long long x) { return x > 0LL ? x : -x; }
void build(int rt, int l, int r) {
  t[rt].l = l;
  t[rt].r = r;
  t[rt].lz = 0;
  t[rt].co = -1;
  if (l == r) {
    t[rt].co = l;
    return;
  }
  int mid = (l + r) >> 1;
  build(rt << 1, l, mid);
  build(rt << 1 | 1, mid + 1, r);
}
void print(int rt) {
  if (t[rt].l == t[rt].r) return;
  print(rt << 1);
  print(rt << 1 | 1);
}
void pushdown(int rt) {
  if (t[rt].l == t[rt].r) return;
  if (t[rt].lz > 0LL) {
    t[rt << 1].val += (t[rt << 1].r - t[rt << 1].l + 1) * t[rt].lz;
    t[rt << 1].lz += t[rt].lz;
    t[rt << 1 | 1].val += (t[rt << 1 | 1].r - t[rt << 1 | 1].l + 1) * t[rt].lz;
    t[rt << 1 | 1].lz += t[rt].lz;
    t[rt].lz = 0LL;
  }
  if (t[rt].co > 0) t[rt << 1].co = t[rt << 1 | 1].co = t[rt].co;
  return;
}
void pushup(int rt) {
  if (t[rt].l == t[rt].r) return;
  if (t[rt << 1].co == t[rt << 1 | 1].co && t[rt << 1].co > 0)
    t[rt].co = t[rt << 1].co;
  else
    t[rt].co = -1;
  t[rt].val = t[rt << 1].val + t[rt << 1 | 1].val;
  return;
}
void insert(int rt, int l, int r, int co) {
  pushdown(rt);
  if (t[rt].l == l && t[rt].r == r && t[rt].co > 0) {
    t[rt].val += 1LL * (t[rt].r - t[rt].l + 1) * abs(co - t[rt].co);
    t[rt].lz += 1LL * abs(co - t[rt].co);
    t[rt].co = co;
    return;
  }
  int mid = (t[rt].l + t[rt].r) >> 1;
  if (r <= mid)
    insert(rt << 1, l, r, co);
  else if (l > mid)
    insert(rt << 1 | 1, l, r, co);
  else {
    insert(rt << 1, l, mid, co);
    insert(rt << 1 | 1, mid + 1, r, co);
  }
  pushup(rt);
}
long long query(int rt, int l, int r) {
  pushdown(rt);
  if (t[rt].l == l && t[rt].r == r) {
    return t[rt].val;
  }
  int mid = (t[rt].l + t[rt].r) >> 1;
  if (r <= mid)
    return query(rt << 1, l, r);
  else if (l > mid)
    return query(rt << 1 | 1, l, r);
  else {
    return query(rt << 1, l, mid) + query(rt << 1 | 1, mid + 1, r);
  }
}
int main() {
  std::ios::sync_with_stdio(false);
  scanf("%d%d", &n, &m);
  build(1, 1, n);
  for (int i = 1; i <= m; i++) {
    int flag;
    scanf("%d", &flag);
    if (flag == 1) {
      scanf("%d%d%d", &l, &r, &x);
      insert(1, l, r, x);
    } else {
      scanf("%d%d", &l, &r);
      cout << query(1, l, r) << endl;
    }
  }
  return 0;
}
