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 = 100100;
const int inf = (int)1e9;
struct node {
  long long sum;
  long long val;
  bool push;
  inline void combine(const node& left, const node& right) {
    sum = left.sum + right.sum;
  }
} T[N << 2];
struct segmentTree {
  int n;
  segmentTree(int m) : n(m) { build(0, 0, n - 1); }
  void build(int x, int l, int r) {
    T[x].val = 0;
    T[x].push = false;
    if (l == r) {
      T[x].sum = 0ll;
      return;
    }
    int m = (l + r) >> 1;
    build((x << 1) + 1, l, m);
    build((x << 1) + 2, m + 1, r);
    T[x].combine(T[(x << 1) + 1], T[(x << 1) + 2]);
  }
  inline void resolve(int x, int l, int r) {
    if (T[x].push) {
      T[x].sum += T[x].val * (r - l + 1);
      if (l != r) {
        T[(x << 1) + 1].val += T[x].val;
        T[(x << 1) + 2].val += T[x].val;
        T[(x << 1) + 1].push = T[(x << 1) + 2].push = true;
      }
      T[x].push = false;
      T[x].val = 0;
    }
  }
  void updateRange(int x, int l, int r, int ql, int qr, long long v) {
    resolve(x, l, r);
    if (l > qr || r < ql) return;
    if (l >= ql && r <= qr) {
      T[x].val += v;
      T[x].push = true;
      resolve(x, l, r);
      return;
    }
    int m = (l + r) >> 1;
    updateRange((x << 1) + 1, l, m, ql, qr, v);
    updateRange((x << 1) + 2, m + 1, r, ql, qr, v);
    T[x].combine(T[(x << 1) + 1], T[(x << 1) + 2]);
  }
  long long queryRange(int x, int l, int r, int ql, int qr) {
    resolve(x, l, r);
    if (l >= ql && r <= qr) return T[x].sum;
    long long ret = 0;
    int m = (l + r) >> 1;
    if (ql <= m) ret += queryRange((x << 1) + 1, l, m, ql, qr);
    if (qr > m) ret += queryRange((x << 1) + 2, m + 1, r, ql, qr);
    return ret;
  }
  void update(int ql, int qr, long long v) {
    updateRange(0, 0, n - 1, ql, qr, v);
  }
  long long query(int ql, int qr) { return queryRange(0, 0, n - 1, ql, qr); }
};
int main() {
  int n, m;
  scanf("%d", &n), scanf("%d", &m);
  set<pair<int, int> > intervals;
  for (int i = 0; i < (n); ++i) intervals.insert(make_pair(i, i + 1));
  intervals.insert(make_pair(n, inf));
  segmentTree st(n);
  for (int z = 0; z < (m); ++z) {
    int type, l, r;
    scanf("%d", &type), scanf("%d", &l), scanf("%d", &r);
    --l, --r;
    if (type == 1) {
      int x;
      scanf("%d", &x);
      set<pair<int, int> >::iterator lt, rt, it;
      lt = intervals.upper_bound(make_pair(l, inf));
      --lt;
      rt = intervals.upper_bound(make_pair(r, inf));
      for (it = lt; it != rt;) {
        int lft = max(it->first, l);
        int col = it->second;
        ++it;
        int rgt = min(it->first - 1, r);
        st.update(lft, rgt, abs(x - col));
      }
      int rStart = it->first;
      --it;
      int lStart = lt->first, lCol = lt->second, rCol = it->second;
      intervals.erase(lt, rt);
      intervals.insert(make_pair(l, x));
      if (rStart != r + 1) intervals.insert(make_pair(r + 1, rCol));
      if (lStart != l) intervals.insert(make_pair(lStart, lCol));
    } else {
      cout << st.query(l, r) << "\n";
    }
  }
  return 0;
}
